Question: The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $3.3$ years; the standard deviation is $0.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living less than $1.9$ years.
$3.3$ $2.6$ $4$ $1.9$ $4.7$ $1.2$ $5.4$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $3.3$ years. We know the standard deviation is $0.7$ years, so one standard deviation below the mean is $2.6$ years and one standard deviation above the mean is $4$ years. Two standard deviations below the mean is $1.9$ years and two standard deviations above the mean is $4.7$ years. Three standard deviations below the mean is $1.2$ years and three standard deviations above the mean is $5.4$ years. We are interested in the probability of a lizard living less than $1.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lizards will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the lizards will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $1.9$ years and the other half $({2.5\%})$ will live longer than $4.7$ years. The probability of a particular lizard living less than $1.9$ years is ${2.5\%}$.